1

Lecture 3Lecture 31.1. The The Debye theoryDebye theory. Gases and polar molecules in . Gases and polar molecules in

non-polar solvent.non-polar solvent.

2.2. The reaction field of a non-polarizable point dipoleThe reaction field of a non-polarizable point dipole

3.3. The internal and the direction fields for polar The internal and the direction fields for polar dielectrics. dielectrics.

4.4. The The Onsager equationOnsager equation. The dielectric properties of . The dielectric properties of polar non-associative liquids.polar non-associative liquids.

2

Debye theory; Gases and Debye theory; Gases and polar molecules in non-polar polar molecules in non-polar solventsolvent

Debye theory; Gases and Debye theory; Gases and polar molecules in non-polar polar molecules in non-polar solventsolvent In Debye theory the internal field EEii and directing field EEdd were

not distinguished from each other. Debye used for both fields the Lorentz's one, given by (2.54):

EE3

2

L

If we will substitute this expression into (2.49)

kdk

kikk

k kTN )(

3)(

4

1 2

EE=E

for both EEii and EEdd, then we can find:

1

2 3

2

=4

3N k

kk

k

kT(3.1)

3

This is generally called the Debye equation. It was the first relationship that made the connection between the molecular molecular parameterparameter of the substance being tested and the phenomenological (macroscopic)phenomenological (macroscopic) parameter that can be experimentally measured.Using the molar polarization [P][P] , defined by (2.57)

[ ]PM

d

1

2

we can write the Debye equationDebye equation for a pure compound as:

[ ]P NkTA

4

3 3

2

(3.2)

Thus, according to the Debye equationDebye equation,, the molar polarization of a compound at a given temperature is constantis constant. It must be independent of the pressure and have the same value in gaseous and the liquid state.

4

kT

N kk

kk 3

412 =

kT

N kk

kk 3

412 = (3.3)

In this form, the Debye equationDebye equation is still used. For gases at

normal pressures the difference between the fields EE, EEii,EEdd,

and EELL can be neglected. For pure compounds equation (3.2)

changes into:

kTN

341

2 = (3.4)

But in many cases, however, the Debye equation is in considerable disagreementdisagreement with the experiment. It works very nice for gases at normal pressuresgases at normal pressures. In this case one has --1<<11<<1 and equation (3.1) can be written as:

5

The reaction fieldThe reaction fieldWhen a molecule with permanent dipole strengthpermanent dipole strength is surrounded by other particles, the inhomogeneous field of the permanent dipole polarizes its environment. In the surrounding particles moments proportional to the polarizability are induced, and if these particles have a permanent dipole moment their orientation is influenced. To calculate this effect one can use a simple model: an ideal an ideal dipole in a center of a spherical cavity. dipole in a center of a spherical cavity. The field of the dipole in such a cavity polarizes the surrounding matter, and the The field of the dipole in such a cavity polarizes the surrounding matter, and the resulting in homogeneous polarization of the environment will give rise to a resulting in homogeneous polarization of the environment will give rise to a field at the dipole, which is called the reaction fieldfield at the dipole, which is called the reaction field R.R. For reasons of symmetry R will have the same direction as the dipole vector , and it is obvious that R will be proportional to as long as no saturation effects occur.

Thus: μR f (3.5)The factor f f is called the factor of reaction field. If the cavity is not spherical, this factor must be replaced by a reaction field reaction field tensortensor.

6

The reaction field of a non-polarizable point dipoleThe reaction field of a non-polarizable point dipoleThe reaction field of a non-polarizable point dipoleThe reaction field of a non-polarizable point dipole

The model consists of a spherical cavity with radius aa in a continuous dielectric of dielectric permittivity . The point dipole with moment is situated in the center of the cavity (fig.3.1). In this very simplified model of the interaction between a permanent dipole moment and its surroundings, the value chosen for aa influences the results.

a

z

a

z

Fig.3.1

Let us assume that only one kind of molecule is presented:

1334 Na (3.6)

It implies that the volume of the cavity is equal to the volume available to each molecule.

To calculate the reaction field we have to know the potential in the cavity due both to the dipole itself and to the interaction of the dipole with the surrounding dielectric.

7

Let the center of the dipole to be the origin of a coordinate system and choose the direction of the z-axis along the dipole vector. It will be the symmetry about the z-axis. As it was shown in the previous lecture the general solution of following relation in the terms of Legendre polynomial represents Laplaces’s equationLaplaces’s equation:

1 10

2 10

A rB

rP

C rD

rP

nn n

nn

n

nn n

nn

n

(cos )

(cos )

where 11 is the potential outside the sphere and 22 is the potential inside the sphere.

The boundary conditions in this case are:

1. 01 r(3.7)

2. arar 21(3.8)

3.arar dr

d

dr

d

21 (3.9)

8

The only source of the field lines within the cavity is the permanent dipole .. The potential due to an ideal dipole along the z-axis is given by:

r 2 cos

thus all coefficients Dn are zero except DD11, which has the value DD11==. Using the bound conditions and the same procedure that we used in previous lecture, we can get

1 2

3

2 1

rcos (3.10)

2 2 3

2 1

r a

rcos cos( )

2 + 1(3.11)

According to (3.11), the field in the cavity is a superposition of the dipole field in vacuum and a uniform field RR, given by:

R

1 2 1

2 13a

( )

(3.12)

Comparing this result with Eqn. (3.4) we conclude that the factor of the reaction field is given by:

kTN

341

2 = (3.4)

9

Formally, the field of dielectric can be described as the field

of a virtual dipole cc at the center of the cavity, given by:

The presented model involves a number of simplifications, since the dipoledipole is assumed to be idealideal and located at the located at the

center of the moleculecenter of the molecule,, which is supposed to be spherical spherical

and surrounded by a continuous dielectric. and surrounded by a continuous dielectric.

fa

1 2 1

2 13

( )

fa

1 2 1

2 13

( ) (3.13)

c 3

2 1

(3.14)

10

The reaction field of a polarized point dipoleThe reaction field of a polarized point dipoleLet us now consider the case of a polarizable permanent dipole, having an average polarizability . In this case the reaction field RR induces a dipole RR and therefore satisfies the equation:

)f( RμR (3.15)

where is the permanent dipole moment. Therefore:

R

f

f1 (3.16)

Eliminating f f by using (3.13) we have:

R

12 1

2 1 3

3

( ) a

a

1 (3.17)

fa

1 2 1

2 13

( )

fa

1 2 1

2 13

( )

11

To give an approximate calculation of the numerical values of RR in the case pure nonassociated polar liquids containing one kind of particle we use approximation (3.6) and the relation:

a

n

n3

2

2

1

2

(3.18)

where nn is the index of refraction. The substitution of (3.6) and (3.18) into (3.17) leads to:

R

4

3

2 1

2

2

32

2

Nn

n( )

(3.19)

The number of particles per cm3 , NN, can be computed from:

Nd

MN A (3.20)

where MM being the molecular weight of the substance, dd density and NNA Avagadro's number. For usual values of (0.5-5 D) the order of magnitude of the reaction field strength is 5-5-50 1050 1066 v/cm. v/cm.

12

Under the influence of the reaction field the dipole momentthe dipole moment is is increased considerablyincreased considerably; the increased moment is:

Rμμ * (3.21)

The combination of (3.21) and (3.16), and taking into account (3.13) gives:

11

2 1

2 13

fa

( )(3.22)

Using (3.18) we can calculate the increase of the dipole moment:

2 1

2

2

32

2 n

n

2 1

2

2

32

2 n

n (3.23)

13

The internal and the direction fields for The internal and the direction fields for polar dielectrics; The Onsager equationpolar dielectrics; The Onsager equationThe internal and the direction fields for The internal and the direction fields for polar dielectrics; The Onsager equationpolar dielectrics; The Onsager equation

In the case of polar dielectrics, the molecules have a

permanent dipole moment , and both parts of the

fundamental equation (equation 2.49 from previous lecture)

must be taken into account.

1

4 3

2

E = E EN kk

k i kk

d kkT( ) ( )

In the case of non-polar liquids the internal field can be considered as the sum of two parts; one being the cavity field and the other the reaction field of the dipole induced in the

molecule EEii=E=Ecc+R.+R.

For polar molecules the internal field can also built up from the cavity field and the reaction field, taking into account now the reaction field of the total dipole moment of the molecule.

14

It means that in a spherical cavity the permanent dipole in a spherical cavity the permanent dipole moment and the reaction field caused by it will have the moment and the reaction field caused by it will have the same direction.same direction. Therefore, this reaction field RR does not influence the direction of the dipole moment of the molecule under consideration, and does not contribute to the directing field EEdd.

REE di- (3.24)

Since the reaction field RR belongs to one particular

orientation of the dipole moment, the difference between EEii

and EEdd will give by the value of the reaction field averaged

over all orientations of the polar molecule:

On the other hand, the reaction field does contribute to the

internal field EEii, because it polarizes the molecule. As a

result, we find a difference between the internal field EEii and

the directing field EEdd.

The angle between the reaction fieldreaction field of the permanent part of the dipole moment and the permanent dipole moment permanent dipole moment

itselfitself will be constant during the movements of the molecule.

15

The direction field EEdd can be obtained by the following procedure: ) remove the permanent dipoleremove the permanent dipole of a molecule without changing its polarizability;

) let the surrounding dielectric adapt itself to the new situation;

) then fixfix the charge distribution of the surroundings and removeremove the central molecule.

The average field in the cavityaverage field in the cavity so obtained is equal to the value of EEdd that is to be calculated, since we have eliminated

the contribution of RR to EEii by removing the permanent dipole removing the permanent dipole

of the moleculeof the molecule.With the aid of this procedure it is possible to calculate EEdd in

a simple way. If in the above procedure we had also to removed the polarizability of the molecule before fixing the charge distribution of the surroundings, we would formed a

physical cavity in the dielectricphysical cavity in the dielectric, i.e. a cavity where the surroundings are allowed to adapt themselves to the new situation.

16

For a spherical cavity, the homogeneous cavity field EEcc is given by eqn (2.14) from the previous lecture:

E EC 3

2 1

To calculate EEdd, the polarizability of the molecule must now be

taken into account. The field EEdd causes a dipole EEdd with

reaction field ffEEdd where f f is the reaction field factor given by

(3.13). Thus, EEdd will be given by the equation:

E E Ed c df (3.25)

or

E Ed cf

1

1 (3.26)

17

Combining (2.14) and (3.26) we have:

If the dielectric consists of different kinds of molecules, we must calculate the directing field for each kind of molecule separately. We then have for the k-th kind of molecule:

with

where is the dielectric constant of the mixture and aakk the

radius of the cavity belong to the particle of the k-thk-th kind.

EE12

3

1

1

fd (3.27)

EE12

3

1

1)( k

kkd f

(3.28)

fak

k

1 2 1

2 13

( ) (3.29)

18

The internal field is now found from eqn.(3.24):

where RR is the total reaction field connected with the permanent part of the dipole moment. This will be given by the expression for the reaction field of the polarizable point dipole, (equation 3.16):

R f

f1

Therefore, <<RR> > is given by:

REE di (3.30)

μR

f

f

1(3.31)

in which <<>> is the value of averaged over all orientations. <<>> is given by equation (2.46) from the previous lecture:

dEμ

2

Hence:

19

E

EEE

12

3

1

1

11

12

2

ff

f

f

fddi =

(3.32)

For a mixture of different kinds of molecules we find for the

internal field at k-thk-th kind of molecule:

EE12

3

1

1

11

2

kk

k

kk

kk ff

f

=i (3.33)

It must be stressed that all these relations are hold only for for

spherical particles.spherical particles. It is not necessary that the dipole be

situated in the center of the sphere, however, if the

polarizability of these particles is considered to be point point

polarizabilitypolarizability in the center of the molecule or homogeneously

distributed over the whole sphere.

20

kdk

kikk

k kTN )(

3)(

4

1 2

EE=E

We can now substitute eqns. (3.28) and (3.33) for the

directing and the internal field, respectively, into the

fundamental equation, eqn. (2.49) from the previous lecture:

We then find:

kkk

kkkk fkTf

N

13

1

1

1

12

1)+1)(2-(= (3.34)

in which ffkk is given by (3.29):

fak

k

1 2 1

2 13

( )

21

Taking for the radius of the cavity and the polarizability the

values of these quantities as found for each pure compound

(eqn. (3.6) and (2.59) from the pervious lecture)

13

kNa3

4

kkkN

3

4

2

1

we obtain for the ratio k

ka3

2)(

1)(3

k

k

k

k

a (3.35)

22

From this we find:

1

1

2 2 1

3 2

f k k

k

k

(3.36)

Hence, (3.34) can be written in the form:

kTN

k

kk

k

k

kk 323

122

2

2

4

1)-(

=(3.37)

For pure dipole liquids it is useful to substitute from the

equation of Clausiuss-Mossotti eqn.(2.59)

kT

N

92

122

2

1

4

3

4

1-2

2

=

kT

N

92

122

2

1

4

3

4

1-2

2

= (3.38)

Hence:

23

2

2

)2(

)2)((

4

9

N

kT2

2

)2(

)2)((

4

9

N

kT(3.39)

This equation is generally called the Onsager Onsager

equationequation. It makes possible the computation of the

permanent dipole moment of a molecule from the

dielectric constant of the pure dipole liquid if the

density and are known.